Question 200: Electrostats & Capacitance

Question

A capacitor is connected to a 20 V battery through a resistance of 10$$\Omega$$. It is found that the potential difference across the capacitor rises to 2 V in 1 $$\mu$$s. The capacitance of the capacitor is __________ $$\mu$$F. Given : $$\ln \left( {{{10} \over 9}} \right) = 0.105$$

Accepted Answer

Given, the peak voltage of the battery, V
0
= 20V
The voltage of the battery, V = 2V
Time, t = 1 $$\mu$$s = 1 $$	imes$$ 10
$$-$$6
s
Resistance of the capacitor, R = 10 $$\Omega$$
As we know that,
V = V
0
(1 $$-$$ e
$$-$$t/RC
)
Substituting the values in the above equation, we get
2 = 20(1 $$-$$ e
$$-$$t/RC
)
$$ \Rightarrow {t \over {RC}} = \ln \left( {{{10} \over 9}} ight)$$
$$ \Rightarrow C = {t \over {R\ln (10/9)}} = {{{{10}^{ - 6}}} \over {10 	imes \ln (10/9)}} = 0.95$$ $$\mu$$F
$$	herefore$$ The capacitance of the capacitor is 0.95 $$\mu$$F.

Answer

9.52

Answer

0.95

Answer

0.105

Answer

1.85

Electrostats & Capacitance

A capacitor is connected to a 20 V battery through a resistance of 10$\Omega. It is found that the potential difference across the capacitor rises to 2 V in 1 \mus. The capacitance of the capacitor is __________ \muF. Given : \ln \left( {{{10} \over 9}} \right) = 0.105

Choose the correct answer: